1–12 ■ PROBLEM-SOLVING TECHNIQUE
The first step in learning any science is to grasp the fundamentals and to gain
a sound knowledge of it. The next step is to master the fundamentals by test-
ing this knowledge. This is done by solving significant real-world problems.
Solving such problems, especially complicated ones, require a systematic
approach. By using a step-by-step approach, an engineer can reduce the solu-
tion of a complicated problem into the solution of a series of simple prob-
lems (Fig. 1–57). When you are solving a problem, we recommend that you
use the following steps zealously as applicable. This will help you avoid
some of the common pitfalls associated with problem solving.
Step 1: Problem Statement
In your own words, briefly state the problem, the key information given,
and the quantities to be found. This is to make sure that you understand the
problem and the objectives before you attempt to solve the problem.
Step 2: Schematic
Draw a realistic sketch of the physical system involved, and list the rele-
vant information on the figure. The sketch does not have to be something
elaborate, but it should resemble the actual system and show the key fea-
tures. Indicate any energy and mass interactions with the surroundings.
Listing the given information on the sketch helps one to see the entire
problem at once. Also, check for properties that remain constant during a
process (such as temperature during an isothermal process), and indicate
them on the sketch.
Chapter 1 | 33
4
3
2
3.5
2.5
1.5
1
0.5
0
0102030
P, kPa
z, m
40 50 60
FIGURE 1–56
The variation of gage pressure with
depth in the gradient zone of the solar
pond.
SOLUTION
PROBLEM
HARD WAY
EASY WAY
FIGURE 1–57
A step-by-step approach can greatly
simplify problem solving.
Performing the integration gives the variation of gage pressure in the gradi-
ent zone to be
Then the pressure at the bottom of the gradient zone (z H 4 m)
becomes
Discussion The variation of gage pressure in the gradient zone with depth is
plotted in Fig. 1–56. The dashed line indicates the hydrostatic pressure for
the case of constant density at 1040 kg/m^3 and is given for reference. Note
that the variation of pressure with depth is not linear when density varies
with depth.
54.0 kPa 1 gage 2
sinh^1 atan
p
4
4
4
ba
1 kN
1000 kg#m>s^2
b
P 2 8.16 kPa 1 1040 kg>m^321 9.81 m>s^22
41 4 m 2
p
PP 1 r 0 g
4 H
p
sinh^1 atan
p
4
z
H
b
SEE TUTORIAL CH. 1, SEC. 12 ON THE DVD.
INTERACTIVE
TUTORIAL